question_answer

Using properties of determinants, prove that

\[\left| \begin{matrix}    {{a}^{2}}+1 & ab & ac  \\    ab & {{b}^{2}}+1 & bc  \\    ca & ca & {{c}^{2}}+1  \\ \end{matrix} \right|=1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}.\]?

OR

Prove that

\[\left| \begin{matrix}    {{(b\,+c)}^{2}} & {{a}^{2}} & {{a}^{2}}  \\    {{b}^{2}} & {{(c\,+a)}^{2}} & {{b}^{2}}  \\    {{c}^{2}} & {{c}^{2}} & {{(a\,+b)}^{2}}  \\ \end{matrix} \right|=2abc{{(a+b+c)}^{3}}.\]

Answer:

Let \[\Delta =\left| \begin{matrix}    {{a}^{2}}+1 & ab & ac  \\    ab & {{b}^{2}}+1 & bc  \\    ca & ca & {{c}^{2}}+1  \\ \end{matrix} \right|\]

On applying \[{{R}_{1}}\to a{{R}_{1}},\] \[{{R}_{2}}\to b{{R}_{2}}\,\,and\,\,{{R}_{3}}\to c{{R}_{3}},\]

We get

\[\Delta =\frac{1}{abc}\left| \begin{matrix}    a({{a}^{2}}+1) & {{a}^{2}}b & {{a}^{2}}c  \\    a{{b}^{2}} & b({{b}^{2}}+1) & {{b}^{2}}c  \\    {{c}^{2}}a & {{c}^{2}}b & c({{c}^{2}}+1)  \\ \end{matrix} \right|\]

On applying \[{{C}_{1}}\to \frac{1}{a}{{C}_{1}},\] \[{{C}_{2}}\to \frac{1}{b}{{C}_{2}}\] and \[{{C}_{3}}\to \frac{1}{c}{{C}_{3}},\] we get

\[\Delta =\frac{abc}{abc}\,\left| \begin{matrix}    {{a}^{2}}+1 & {{a}^{2}} & {{a}^{2}}  \\    {{b}^{2}} & {{b}^{2}}+1 & {{b}^{2}}  \\    {{c}^{2}} & {{c}^{2}} & {{c}^{2}}+1  \\ \end{matrix} \right|\]

On applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}},\] we get

\[\Delta =\left| \begin{matrix}    1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}  \\    {{b}^{2}} & {{b}^{2}}+1 & {{b}^{2}}  \\    {{c}^{2}} & {{c}^{2}} & {{c}^{2}}+1  \\ \end{matrix} \right|\]On taking common \[(1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}})\] from \[{{R}_{1}},\]we get

\[\Delta =(1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}})\left| \begin{matrix}    1 & 1 & 1  \\    {{b}^{2}} & {{b}^{2}}+1 & {{b}^{2}}  \\    {{c}^{2}} & {{c}^{2}} & {{c}^{2}}+1  \\ \end{matrix} \right|\]

On applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}},\] we get

\[\Delta =(1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}})\left| \begin{matrix}    1 & 0 & 0  \\    {{b}^{2}} & 1 & 0  \\    {{c}^{2}} & 0 & 1  \\ \end{matrix} \right|\]

Now expanding along \[{{R}_{1}},\] we get

\[\Delta =(1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}})\,1\,[1-0]\]

\[=1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]              Hence proved.

OR

Let the given determinant be \[\Delta .\] then,

\[\Delta =\left| \begin{matrix}    {{(b\,+c)}^{2}} & {{a}^{2}} & {{a}^{2}}  \\    {{b}^{2}} & {{(c\,+a)}^{2}} & {{b}^{2}}  \\    {{c}^{2}} & {{c}^{2}} & {{(a\,+b)}^{2}}  \\ \end{matrix} \right|\]

\[=\left| \begin{matrix}    {{(b\,+c)}^{2}}-{{a}^{2}} & 0 & {{a}^{2}}  \\    0 & {{(c\,+a)}^{2}}-{{b}^{2}} & {{b}^{2}}  \\    {{c}^{2}}-{{(a\,+b)}^{2}} & {{c}^{2}}-{{(a\,+b)}^{2}} & {{(a\,+b)}^{2}}  \\ \end{matrix} \right|\]

\[[{{C}_{1}}\to {{C}_{1}}-{{C}_{3}}\,\,\,and\,\,\,{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}]\]

\[=\left| \begin{matrix}    (a+b+c)(b+c-a) & 0 & {{a}^{2}}  \\    0 & (a+b+c)(c+a-b) & {{b}^{2}}  \\    (a+b+c)(c-a-b) & (a+b+c)(c-a-b) & {{(a+b)}^{2}}  \\ \end{matrix} \right|\]\[={{(a\,+b\,+c)}^{2}}\cdot \left| \begin{matrix}    (b\,+c\,-a) & 0 & {{a}^{2}}  \\    0 & c+a-b & {{b}^{2}}  \\    -\,c-a-b & -\,c-a-b & 2ab  \\ \end{matrix} \right|\]

[taking \[(a\,+b\,+c)\]common from \[{{C}_{1}}\]and \[{{C}_{2}}\]both]

\[={{(a\,+b\,+c)}^{2}}\cdot \left| \begin{matrix}    (b\,+c\,-a) & 0 & {{a}^{2}}  \\    0 & c+a-b & {{b}^{2}}  \\    -\,2b & -\,2a & 2ab  \\ \end{matrix} \right|\]

\[[{{R}_{3}}\to {{R}_{3}}-({{R}_{1}}+{{R}_{2}})]\]

\[={{(a\,+b+c)}^{2}}\,[(b\,+c\,-a)\{(c\,+a-b)\cdot 2ab\,+2a{{b}^{2}}\}\]

\[+{{a}^{2}}\{0+2b(c\,+a\,-b)\}]\]

\[={{(a\,+b\,+c)}^{2}}[b\,+c\,-a]\cdot 2ab\{(c\,+a\,-b\,+b)\}\]

\[+\,2{{a}^{2}}b(c\,+a\,-b)]\]

\[=2ab{{(a\,+b\,+c)}^{2}}\{(b\,+c\,-a)\,(c\,+a)+a(c+a-b)\}\]

\[=2ab{{(a\,+b\,+c)}^{2}}\cdot \{bc\,+ab+{{c}^{2}}+ac-ac-{{a}^{2}}\]

\[+ac+{{a}^{2}}-ab\}\]

\[=2ab\,{{(a\,+b\,+c)}^{2}}\{bc\,+{{c}^{2}}+ac\}\]

\[=2abc\,{{(a\,+b\,+c)}^{3}}.\]

Hence, \[\Delta =2abc\,{{(a\,+b\,+c)}^{3}}.\]

Hence proved.